Đề kiểm tra 1 tiết chương 1: Hàm số lượng giác và phương trình lượng giác - Đề số 2

\(D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{4} + k\dfrac{\pi }{3},\dfrac{{n\pi }}{5};k,n \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{5} + k\dfrac{\pi }{3},\dfrac{{n\pi }}{5};k,n \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{6} + k\dfrac{\pi }{4},\dfrac{{n\pi }}{5};k,n \in \mathbb{Z}} \right\}\)

\(x = \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)\)

\(x =  - \dfrac{\pi }{3} + k2\pi \left( {k \in Z} \right)\)          

\(x = \dfrac{\pi }{6} + k\pi \left( {k \in Z} \right)\)

\(x = \pi \).

\(x = \dfrac{\pi }{3}\).

\(x = \dfrac{\pi }{2}\).

$k2\pi \left( {k \in Z} \right)$

$k2\pi ;\dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$

$k\pi ;\dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$

\(y = \sin x\)

\(y = \cos x\)

\(y = \sin 2x\)  

\(D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{3} + k\dfrac{\pi }{2};k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{4} + k\dfrac{\pi }{2};k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{8} + k\dfrac{\pi }{2};k \in \mathbb{Z}} \right\}\)

\(y = \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)\)

\(x = k\pi \left( {k \in Z} \right)\)

\(y = \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\)

\(\left( {\dfrac{\pi }{2} + k2\pi ;\dfrac{{3\pi }}{2} + k2\pi } \right)\)

\(\left( { - \pi  + k2\pi ;k2\pi } \right)\)

\(R\)

$x =  - \dfrac{\pi }{2} + k2\pi $.

$x = \dfrac{\pi }{2} + k\pi $.

$x = k\pi $.

\(m =  - 3\)

\(m =  - 2\)      

\(m = 3\)

\(x = \pi  + k2\pi ,x = \dfrac{\pi }{4} + k\pi \).

\(x = \pi  + k2\pi ,x =  - \dfrac{\pi }{4} + k2\pi \).

\(x = \pi  + k2\pi ,x = \dfrac{\pi }{4} + k2\pi \).

\(\left[ \begin{array}{l}x = \alpha  + k\pi \\x = \pi  - \alpha  + k\pi \end{array} \right.\left( {k \in Z} \right)\)    

\(\left[ \begin{array}{l}x = \alpha  + k2\pi \\x =  - \alpha  + k2\pi \end{array} \right.\left( {k \in Z} \right)\)      

\(\left[ \begin{array}{l}x = \alpha  + k\pi \\x =  - \alpha  + k\pi \end{array} \right.\left( {k \in Z} \right)\)

\(\left( { - 1;1} \right)\)           

\(R\)

\(\left[ {0;1} \right]\)

\({T_0} = \dfrac{\pi }{2}\)

${T_0} = \pi $

\({T_0} = \dfrac{\pi }{4}\)

$\left[ \begin{array}{l}x = k\pi \\x = \dfrac{\pi }{3} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)$           

$x = k\pi \,\,\,\left( {k \in \mathbb{Z}} \right)$

$\left[ \begin{array}{l}x = k2\pi \\x =  - \dfrac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)$

$x = \dfrac{\pi }{3} + k2\pi $.

$x = \dfrac{\pi }{6} + k\pi $.

$x = k\pi $.

\(\max y = 6;\min y =  - 2\)

\(\max y = 4;\min y =  - 4\)

\(\max y = 6;\min y =  - 1\)

\(R\backslash \left\{ {\dfrac{\pi }{2} + k\pi ,k \in Z} \right\}\)

\(R\backslash \left\{ {k\pi ,k \in Z} \right\}\)

\(R\)

\(\left| a \right| \ge 1\)

\(\left| a \right| = 1\)

\(\left| a \right| \ne 1\)

\(m \ne \dfrac{1}{2}\)

\(m = \dfrac{1}{2}\)

\(\dfrac{1}{3} < m < 1\)

\(x = \arctan \dfrac{{1 + \sqrt {73} }}{{12}} + \dfrac{{k\pi }}{2}\,\,,\,\,x = \arctan \dfrac{{1 - \sqrt {73} }}{{12}} + \dfrac{{k\pi }}{2}\,\,\left( {k \in Z} \right)\)

\(x = \dfrac{1}{2}\arctan \dfrac{{1 + \sqrt {73} }}{6} + \dfrac{{k\pi }}{2}\,\,,\,\,x = \dfrac{1}{2}\arctan \dfrac{{1 - \sqrt {73} }}{6} + \dfrac{{k\pi }}{2}\,\,\left( {k \in Z} \right)\)

\(x = \arctan \dfrac{3}{2} + \dfrac{{k\pi }}{2}\,\,,\,\,x = \arctan \left( { - 1} \right) + \dfrac{{k\pi }}{2}\,\,\left( {k \in Z} \right)\)

\(\left\{ \begin{array}{l}x = \pi  + k\pi \\y =  - \dfrac{\pi }{3} - k\pi \end{array} \right.\).

\(\left\{ \begin{array}{l}x = \dfrac{{2\pi }}{3} + k\pi \\y =  - k\pi \end{array} \right.\).

\(\left\{ \begin{array}{l}x = \dfrac{{5\pi }}{6} + k\pi \\y =  - \dfrac{\pi }{6} - k\pi \end{array} \right.\).

\(m > 0\)

\(m < 0\)

\(m \le 1\)

\(0\)     

\(1\)

\(3\)

\(x =  - \dfrac{\pi }{6} + k2\pi \,\,\,\left( {k \in Z} \right)\)

\(x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\,\,\,\left( {k \in Z} \right)\)

\(x =  \pm \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\,\,\,\left( {k \in Z} \right)\)